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\title{\heiti\zihao{2} 习题15.4}
\author{中书君}
\date{\today}
\begin{document}
\maketitle
\section{求下列方程所确定的隐函数 $z=z(x, y)$ 的偏导数 $\dfrac{\partial z}{\partial x}, \dfrac{\partial z}{\partial y}$}
\subsection{$x+y+z=e^{-(x+y+z)}$}
\textbf{解}\quad
记$F(\boldsymbol{x})=e^{-(x+y+z)}-(x+y+z)$
$$
	\begin{aligned}
		\dfrac{\partial F}{\partial x} & =-e^{-(x+y+z)}-1 \\
		\dfrac{\partial F}{\partial y} & =-e^{-(x+y+z)}-1 \\
		\dfrac{\partial F}{\partial z} & =-e^{-(x+y+z)}-1 \\
	\end{aligned}
$$
则由于其有各个连续偏导数,且$F_z\neq 0$,故
$$
	\begin{aligned}
		\dfrac{\partial z}{\partial x} & =-\dfrac{F_x}{F_z}=-1 \\
		\dfrac{\partial z}{\partial y} & =-\dfrac{F_y}{F_z}=-1 \\
	\end{aligned}
$$

\subsection{$x^{3}+y^{3}+z^{3}-3 a x y z=0$}
\textbf{解}\quad
记$F(\boldsymbol{x})=x^{3}+y^{3}+z^{3}-3 a x y z$
$$
	\begin{aligned}
		\dfrac{\partial F}{\partial x} & =3x^2-3ayz \\
		\dfrac{\partial F}{\partial y} & =3y^2-3azx \\
		\dfrac{\partial F}{\partial z} & =3z^2-3axy \\
	\end{aligned}
$$
则由于其有各个连续偏导数,且$F_z\neq 0$,故
$$
	\begin{aligned}
		\dfrac{\partial z}{\partial x} & =-\dfrac{F_x}{F_z}=-\dfrac{x^2-ayz}{z^2-axy} \\
		\dfrac{\partial z}{\partial y} & =-\dfrac{F_y}{F_z}=-\dfrac{y^2-azx}{z^2-axy} \\
	\end{aligned}
$$

\subsection{$\mathrm{e}^{x}-x y z=0$}
\textbf{解}\quad
记$F(\boldsymbol{x})=\mathrm{e}^{x}-x y z$
$$
	\begin{aligned}
		\dfrac{\partial F}{\partial x} & =e^x-yz \\
		\dfrac{\partial F}{\partial y} & =-xz    \\
		\dfrac{\partial F}{\partial z} & =-xy    \\
	\end{aligned}
$$
则由于其有各个连续偏导数,且$F_z\neq 0$,故
$$
	\begin{aligned}
		\dfrac{\partial z}{\partial x} & =-\dfrac{F_x}{F_z}=\dfrac{e^x-yz}{xy} \\
		\dfrac{\partial z}{\partial y} & =-\dfrac{F_y}{F_z}=-\dfrac{z}{y}      \\
	\end{aligned}
$$

\section{设 $\boldsymbol{F}\left(x+y+z, x^{2}+y^{2}+z^{2}\right)=0$, 求 $\dfrac{\partial z}{\partial x}, \dfrac{\partial z}{\partial y}$.}
\textbf{解}\quad
分别对$x,y$求偏导得
$$
	\left\{\begin{array}{l}
		F_1\left(1+\dfrac{\partial z}{\partial x}\right)+F_2\left(2x+2z\cdot\dfrac{\partial z}{\partial x}\right)=0 \\
		F_1\left(1+\dfrac{\partial z}{\partial y}\right)+F_2\left(2y+2z\cdot\dfrac{\partial z}{\partial y}\right)=0
	\end{array}\right.
$$
解得
$$
	\begin{aligned}
		\dfrac{\partial z}{\partial x} & =-\dfrac{F_1+2xF_2}{F_1+2zF_2} \\
		\dfrac{\partial z}{\partial y} & =-\dfrac{F_1+2yF_2}{F_1+2zF_2}
	\end{aligned}
$$

\section{设 $F(x y, y+z, x z)=0$, 求 $\dfrac{\partial z}{\partial x}, \dfrac{\partial^{2} z}{\partial x \partial y}$.}
\textbf{解}\quad
分别对$x,y$求偏导得:
$$
	\begin{aligned}
		\left\{\begin{array}{l}
			F_1\cdot y+F_2 \cdot\dfrac{\partial z}{\partial x}+F_3\cdot\left(z+x\dfrac{\partial z}{\partial x}\right)=0 \\
			F_1\cdot x+F_2\cdot\left(1+\dfrac{\partial z}{\partial y}\right)+F_3\cdot x\dfrac{\partial z}{\partial y}=0
		\end{array}\right.
	\end{aligned}
$$
解得
$$
	\begin{aligned}
		\dfrac{\partial z}{\partial x} & =-\dfrac{F_1y+F_3z}{F_2+F_3x} \\
		\dfrac{\partial z}{\partial y} & =-\dfrac{F_1x+F_2}{F_3x+F_2}
	\end{aligned}
$$
同理,
\begin{equation}
	\begin{aligned}
		\dfrac{\partial^{2} z}{\partial x \partial y} & =\dfrac{\partial }{\partial x}\left(-\dfrac{F_1x+F_2}{F_3x+F_2}\right)                                                                                    \\
		                                              & =-\dfrac{\dfrac{\partial}{\partial x}\left(F_1x+F_2\right)\cdot(F_3x+F_2)-\dfrac{\partial}{\partial x}\left(F_3x+F_2\right)\cdot(F_1x+F_2)}{(F_3x+F_2)^2}
	\end{aligned}
\end{equation}
其中
$$
	\begin{aligned}
		\dfrac{\partial}{\partial x}\left(F_1x+F_2\right) = & F_{11}xy+F_{12}x\dfrac{\partial z}{\partial x}+F_{13}x\left(z+x\dfrac{\partial z}{\partial x}\right)+F_1 \\
		                                                    & +F_{21}y+F_{22}\dfrac{\partial z}{\partial x}+F_{23}\left(z+\dfrac{\partial z}{\partial x}\right)        \\
		\dfrac{\partial}{\partial x}\left(F_3x+F_2\right) = & F_{31}xy+F_{32}x\dfrac{\partial z}{\partial x}+F_{33}x\left(z+x\dfrac{\partial z}{\partial x}\right)+F_3 \\
		                                                    & +F_{21}y+F_{22}\dfrac{\partial z}{\partial x}+F_{23}\left(z+\dfrac{\partial z}{\partial x}\right)
	\end{aligned}
$$
将$\dfrac{\partial z}{\partial x}$带入上式,然后将上式代入(1),即可得
$$
	\begin{aligned}
		\dfrac{\partial^2z}{\partial x\partial y} = & -\dfrac{F_{11}xy-F_12x\dfrac{F_1y+F_3z}{F_2+F_3x}+F_{13}x\left(z-x\dfrac{F_1y+F_3z}{F_2+F_3x}\right)+F_1}{F_3x+F_2}                             \\
		                                            & +\dfrac{F_{21}y-F_{22}\dfrac{F_1y+F_3z}{F_2+F_3x}+F_{23}\left(z-\dfrac{F_1y+F_3z}{F_2+F_3x}\right)}{F_3x+F_2}                                   \\
		                                            & -\dfrac{\left(F_{31}xy-F_{32}\dfrac{F_1y+F_3z}{F_2+F_3x}+F_{33}x\left(z-x\dfrac{F_1y+F_3z}{F_2+F_3x}\right)+F_3\right)(F_1x+F_2)}{(F_3x+F_2)^2} \\
		                                            & +\dfrac{\left(F_{21}y-F_{22}\dfrac{F_1y+F_3z}{F_2+F_3x}+F_{23}\left(z-\dfrac{F_1y+F_3z}{F_2+F_3x}\right)\right)(F_1x+F_2)}{(F_3x+F_2)^2}
	\end{aligned}
$$

\section{设方程组$$\left\{\begin{array}{l}u+v+w+x+y=a \\u^{2}+v^{2}+w^{2}+x^{2}+y^{2}=b^{2} \\u^{3}+v^{3}+w^{3}+x^{3}+y^{3}=c^{3}\end{array}\right.$$确定 $u, v, w$ 为 $x, y$ 的隐函数,求 $u, v, w$ 关于 $x, y$ 的偏导数.}
\textbf{解}\quad
$a,b,c$是常数,所以关于$u,v,w$的Jacobian矩阵为$\left[\begin{array}{ccc}
			1    & 1    & 1    \\
			2u   & 2v   & 2w   \\
			3u^2 & 3v^2 & 3w^2
		\end{array}\right]$从而
$$
	\begin{aligned}
		\dfrac{\partial u}{\partial x} & =-\dfrac{\left|\begin{array}{ccc}
				1    & 1    & 1    \\
				2x   & 2v   & 2w   \\
				3x^2 & 3v^2 & 3w^2
			\end{array}
			\right|}{\left|\begin{array}{ccc}
				1    & 1    & 1    \\
				2u   & 2v   & 2w   \\
				3u^2 & 3v^2 & 3w^2
			\end{array}\right|}=-\dfrac{\left|\begin{array}{ccc}
				1   & 1   & 1   \\
				x   & v   & w   \\
				x^2 & v^2 & w^2
			\end{array}
		\right|}{\left|\begin{array}{ccc}
				1   & 1   & 1   \\
				u   & v   & w   \\
				u^2 & v^2 & w^2
			\end{array}\right|}                                                      \\
		                               & =\dfrac{vw^2-v^2w-xw^2+x^2w+xv^2-x^2v}{vw^2-v^2w-uw^2+u^2w+uv^2-u^2v} \\
		\dfrac{\partial v}{\partial x} & =-\dfrac{\left|\begin{array}{ccc}
				1    & 1    & 1    \\
				2u   & 2x   & 2w   \\
				3u^2 & 3x^2 & 3w^2
			\end{array}
			\right|}{\left|\begin{array}{ccc}
				1    & 1    & 1    \\
				2u   & 2v   & 2w   \\
				3u^2 & 3v^2 & 3w^2
			\end{array}\right|}=-\dfrac{\left|\begin{array}{ccc}
				1   & 1   & 1   \\
				u   & x   & w   \\
				u^2 & x^2 & w^2
			\end{array}
		\right|}{\left|\begin{array}{ccc}
				1   & 1   & 1   \\
				u   & v   & w   \\
				u^2 & v^2 & w^2
			\end{array}\right|}                                                      \\
		                               & =\dfrac{xw^2-x^2w-uw^2+u^2w+ux^2-u^2x}{vw^2-v^2w-uw^2+u^2w+uv^2-u^2v} \\
		\dfrac{\partial w}{\partial x} & =-\dfrac{\left|\begin{array}{ccc}
				1    & 1    & 1    \\
				2u   & 2v   & 2x   \\
				3u^2 & 3v^2 & 3x^2
			\end{array}
			\right|}{\left|\begin{array}{ccc}
				1    & 1    & 1    \\
				2u   & 2v   & 2w   \\
				3u^2 & 3v^2 & 3w^2
			\end{array}\right|}=-\dfrac{\left|\begin{array}{ccc}
				1   & 1   & 1   \\
				u   & v   & x   \\
				u^2 & v^2 & x^2
			\end{array}
		\right|}{\left|\begin{array}{ccc}
				1   & 1   & 1   \\
				u   & v   & w   \\
				u^2 & v^2 & w^2
			\end{array}\right|}                                                      \\
		                               & =\dfrac{vx^2-v^2x-ux^2+u^2x+uv^2-u^2v}{vw^2-v^2w-uw^2+u^2w+uv^2-u^2v}
	\end{aligned}
$$
\clearpage
$$
	\begin{aligned}
		\dfrac{\partial u}{\partial y} & =-\dfrac{\left|\begin{array}{ccc}
				1    & 1    & 1    \\
				2y   & 2v   & 2w   \\
				3y^2 & 3v^2 & 3w^2
			\end{array}
			\right|}{\left|\begin{array}{ccc}
				1    & 1    & 1    \\
				2u   & 2v   & 2w   \\
				3u^2 & 3v^2 & 3w^2
			\end{array}\right|}=-\dfrac{\left|\begin{array}{ccc}
				1   & 1   & 1   \\
				y   & v   & w   \\
				y^2 & v^2 & w^2
			\end{array}
		\right|}{\left|\begin{array}{ccc}
				1   & 1   & 1   \\
				u   & v   & w   \\
				u^2 & v^2 & w^2
			\end{array}\right|}                                                      \\
		                               & =\dfrac{vw^2-v^2w-yw^2+y^2w+yv^2-y^2v}{vw^2-v^2w-uw^2+u^2w+uv^2-u^2v} \\
		\dfrac{\partial v}{\partial y} & =-\dfrac{\left|\begin{array}{ccc}
				1    & 1    & 1    \\
				2u   & 2y   & 2w   \\
				3u^2 & 3y^2 & 3w^2
			\end{array}
			\right|}{\left|\begin{array}{ccc}
				1    & 1    & 1    \\
				2u   & 2v   & 2w   \\
				3u^2 & 3v^2 & 3w^2
			\end{array}\right|}=-\dfrac{\left|\begin{array}{ccc}
				1   & 1   & 1   \\
				u   & y   & w   \\
				u^2 & y^2 & w^2
			\end{array}
		\right|}{\left|\begin{array}{ccc}
				1   & 1   & 1   \\
				u   & v   & w   \\
				u^2 & v^2 & w^2
			\end{array}\right|}                                                      \\
		                               & =\dfrac{yw^2-y^2w-uw^2+u^2w+uy^2-u^2y}{vw^2-v^2w-uw^2+u^2w+uv^2-u^2v} \\
		\dfrac{\partial w}{\partial y} & =-\dfrac{\left|\begin{array}{ccc}
				1    & 1    & 1    \\
				2u   & 2v   & 2y   \\
				3u^2 & 3v^2 & 3y^2
			\end{array}
			\right|}{\left|\begin{array}{ccc}
				1    & 1    & 1    \\
				2u   & 2v   & 2w   \\
				3u^2 & 3v^2 & 3w^2
			\end{array}\right|}=-\dfrac{\left|\begin{array}{ccc}
				1   & 1   & 1   \\
				u   & v   & y   \\
				u^2 & v^2 & y^2
			\end{array}
		\right|}{\left|\begin{array}{ccc}
				1   & 1   & 1   \\
				u   & v   & w   \\
				u^2 & v^2 & w^2
			\end{array}\right|}                                                      \\
		                               & =\dfrac{vy^2-v^2y-uy^2+u^2y+uv^2-u^2v}{vw^2-v^2w-uw^2+u^2w+uv^2-u^2v}
	\end{aligned}
$$

\section{}
\subsection{设 $\left\{\begin{array}{l}u=f(x u, v+y), \\ v=g\left(u-x, v^{2} y\right),\end{array}\right.$ 其中 $f, g$ 都具有一阶连续偏导数,求 $\dfrac{\partial u}{\partial y}$ 和 $\dfrac{\partial v}{\partial y}$.}
\textbf{解}\quad
由于$f, g$ 都具有一阶连续偏导数,所以
$$
	\begin{aligned}
		\dfrac{\partial u}{\partial y} & =f_1x\dfrac{\partial u}{\partial y}+f_2\left(1+\dfrac{\partial v}{\partial y}\right)     \\
		\dfrac{\partial v}{\partial y} & =g_1\dfrac{\partial u}{\partial y}+g_2\left(2vy\dfrac{\partial v}{\partial y}+v^2\right)
	\end{aligned}
$$
解得$\dfrac{\partial u}{\partial y}=-\dfrac{f_{2}\left[\left(2 g_{2} v y-1\right)-v^{2} g_{2}\right]}{\left(2 v y g_{2}-1\right)\left(x f_{1}-1\right)-f_{2} g_{1}},\dfrac{\partial v}{\partial y}=-\dfrac{v^{2}\left(x f_{1}-1\right) g_{2}-f_{2} g_{1}}{\left(2 v y g_{2}-1\right)\left(x f_{1}-1\right)-f_{2} g_{1}}$

\subsection{设 $x u-y v=0, y u+x v=1$, 求 $\dfrac{\partial u}{\partial x}, \dfrac{\partial v}{\partial x}, \dfrac{\partial u}{\partial y}$ 和 $\dfrac{\partial v}{\partial y}$.}
\textbf{解}\quad
此处的$u,v,x,y$不同于上一题.\par
$\boldsymbol{F}(x,y,u,v)$关于$u,v$的Jacobian矩阵为$\left[\begin{array}{cc}
			x & -y \\
			y & x
		\end{array}\right]$.故
$$
	\begin{aligned}
		\left[\begin{array}{cc}
				\dfrac{\partial u}{\partial x} & \dfrac{\partial u}{\partial y} \\
				\dfrac{\partial v}{\partial x} & \dfrac{\partial v}{\partial y}
			\end{array}\right] & =-\left[\begin{array}{cc}
				x & -y \\
				y & x
			\end{array}\right]^{-1}\cdot\left[\begin{array}{cc}
				u & -v \\
				v & u
			\end{array}\right] \\
		                                        & =-\left[\begin{array}{cc}
				\dfrac{x}{x^2+y^2}  & \dfrac{y}{x^2+y^2} \\
				-\dfrac{y}{x^2+y^2} & \dfrac{x}{x^2+y^2}
			\end{array}\right]\cdot\left[\begin{array}{cc}
				u & -v \\
				v & u
			\end{array}\right]
	\end{aligned}
$$
解得
$$
	\begin{aligned}
		\dfrac{\partial u}{\partial x} & =-\dfrac{ux+vy}{x^2+y^2}  \\
		\dfrac{\partial u}{\partial y} & =-\dfrac{-vx+uy}{x^2+y^2} \\
		\dfrac{\partial v}{\partial x} & =-\dfrac{-uy+vx}{x^2+y^2} \\
		\dfrac{\partial v}{\partial y} & =-\dfrac{vy+ux}{x^2+y^2}
	\end{aligned}
$$

\section{设 $u=u(x, y), x=r \cos \theta, y=r \sin \theta$, 证明:$$\dfrac{\partial^{2} u}{\partial x^{2}}+\dfrac{\partial^{2} u}{\partial y^{2}}=\dfrac{\partial^{2} u}{\partial r^{2}}+\dfrac{1}{r} \dfrac{\partial u}{\partial r}+\dfrac{1}{r^{2}} \dfrac{\partial^{2} u}{\partial \theta^{2}}$$}
\begin{proof}
	显然$r=\sqrt{x^{2}+y^{2}},\theta=\arctan \dfrac{y}{x} .$ \par
	设 $u(x, y)=F(r, \theta)$, 则
	$$
		\begin{aligned}
			\dfrac{\partial u}{\partial x} & =\dfrac{\partial u}{\partial r} \cdot \dfrac{\partial r}{\partial x}+\dfrac{\partial u}{\partial \theta} \cdot \dfrac{\partial \theta}{\partial x} \\
			                               & =\dfrac{\partial u}{\partial r} \cdot \dfrac{x}{r}-\dfrac{\partial u}{\partial \theta} \cdot \dfrac{y}{r^{2}}                                      \\
			                               & =\dfrac{\partial u}{\partial r} \cos \theta-\dfrac{\partial u}{\partial \theta} \dfrac{\sin \theta}{r}
		\end{aligned}
	$$
	同理可得
	$$
		\dfrac{\partial u}{\partial y}=\dfrac{\partial u}{\partial r} \sin \theta+\dfrac{\partial u}{\partial \theta}\dfrac{\cos \theta}{r}
	$$
	又因为
	$$
		\begin{aligned}
			\dfrac{\partial^{2} u}{\partial x^{2}} & =\dfrac{\partial}{\partial r}\left(\dfrac{\partial u}{\partial x}\right) \cdot \dfrac{\partial r}{\partial x}+\dfrac{\partial}{\partial \theta}\left(\dfrac{\partial u}{\partial x}\right) \cdot \dfrac{\partial \theta}{\partial x}                                                                                                                            \\
			                                       & =\dfrac{\partial}{\partial r}\left(\dfrac{\partial u}{\partial r} \cos \theta-\dfrac{\partial u}{\partial \theta} \dfrac{\sin \theta}{r}\right) \cos \theta-\dfrac{\partial}{\partial \theta}\left(\dfrac{\partial u}{\partial r} \cos \theta-\dfrac{\partial u}{\partial \theta} \dfrac{\sin \theta}{r}\right) \cdot \dfrac{\sin \theta}{r}                    \\
			                                       & =\dfrac{\partial^{2} u}{\partial r^{2}} \cos ^{2} \theta-2 \dfrac{\partial^{2} u}{\partial r \partial \theta} \dfrac{\sin \theta \cos \theta}{r}+\dfrac{\partial^{2} u}{\partial \theta^{2}} \dfrac{\sin ^{2} \theta}{r^{2}}+2 \dfrac{\partial u}{\partial r} \dfrac{\sin \theta \cos \theta}{r^{2}}+\dfrac{\partial u}{\partial r} \dfrac{\sin ^{2} \theta}{r}
		\end{aligned}
	$$
	同理可得
	$$
		\dfrac{\partial^{2} u}{\partial y^{2}}=\dfrac{\partial^{2} u}{\partial r^{2}} \sin ^{2} \theta+2 \dfrac{\partial^{2} u}{\partial r \partial \theta} \dfrac{\sin \theta \cos \theta}{r}+\dfrac{\partial^{2} u}{\partial \theta^{2}} \dfrac{\cos ^{2} \theta}{r^{2}}-2 \dfrac{\partial u}{\partial \theta} \dfrac{\sin \theta \cos \theta}{r^{2}}+\dfrac{\partial u}{\partial r} \dfrac{\cos ^{2} \theta}{r}
	$$
	两式相加, 得
	$$
		\begin{aligned}
			\dfrac{\partial^{2} u}{\partial x^{2}}+\dfrac{\partial^{2} u}{\partial y^{2}} & =\dfrac{\partial^{2} u}{\partial r^{2}}+\dfrac{1}{r} \dfrac{\partial u}{\partial r}+\dfrac{1}{r^{2}} \dfrac{\partial^{2} u}{\partial \theta^{2}}      \\
			                                                                              & =\dfrac{1}{r^{2}}\left[r \dfrac{\partial}{\partial r}\left(r \dfrac{\partial u}{\partial r}\right)+\dfrac{\partial^{2} u}{\partial \theta^{2}}\right] \\
			                                                                              & =\dfrac{\partial^{2} u}{\partial r^{2}}+\dfrac{1}{r} \dfrac{\partial u}{\partial r}+\dfrac{1}{r^{2}} \dfrac{\partial^{2} u}{\partial \theta^{2}}
		\end{aligned}
	$$
\end{proof}



\end{document}